A 3 kg object has a velocity (6i-2j) m/s. A) what is its kinetic energy at this time ?
B) find the total work done on the object if its velocity changes to(8i+4j) m/s.
To solve this problem, we need to calculate the kinetic energy of the object at its initial velocity and then find the total work done on the object when its velocity changes.
A) To find the kinetic energy (KE) at the initial velocity, we can use the equation:
KE = (1/2) * mass * velocity^2
Given:
Mass (m) = 3 kg
Initial velocity (v_initial) = 6i - 2j m/s
First, we need to find the magnitude of the velocity vector. The magnitude of a vector v is given by:
|v| = sqrt(vx^2 + vy^2),
where vx and vy are the x and y components of the vector v, respectively.
|velocity| = sqrt((6)^2 + (-2)^2) = sqrt(36 + 4) = sqrt(40) = 2 * sqrt(10)
Now, we can find the kinetic energy:
KE = (1/2) * mass * velocity^2
= (1/2) * 3 kg * (2 * sqrt(10))^2
= (1/2) * 3 kg * (4 * 10)
= (1/2) * 3 kg * 40
= 60 J
Therefore, the kinetic energy of the object at this time is 60 Joules.
B) To find the total work done on the object when its velocity changes, we can use the work-energy principle. The work done on an object is equal to its change in kinetic energy.
Given:
New velocity (v_final) = 8i + 4j m/s
First, we need to find the change in kinetic energy (ΔKE):
ΔKE = KE_final - KE_initial
We already know the initial KE from part A as 60 J. Now, let's find the final kinetic energy (KE_final) using the same formula as above:
|v_final| = sqrt((8)^2 + (4)^2) = sqrt(64 + 16) = sqrt(80) = 4 * sqrt(5)
KE_final = (1/2) * mass * velocity^2
= (1/2) * 3 kg * (4 * sqrt(5))^2
= 6 * 20 * 5
= 600 J
Now, we can find the change in kinetic energy:
ΔKE = KE_final - KE_initial
= 600 J - 60 J
= 540 J
Therefore, the total work done on the object when its velocity changes to (8i + 4j) m/s is 540 Joules.